Let $$ \sum_{n=0}^\infty c_n (z-a)^n $$ be a power series. If the value $$ r=\underset{n\to\infty}{\lim}\left|\frac{c_n}{c_{n+1}}\right| $$ exists (the limit exists and is a real number), it is the radius of convergence $R$ of the power series. This means that the series converges for all $z$ with $|z-a|<R$.
- Is the radius of convergence $R$ the whole $\mathbb{R}$ if the above limit $r$ is $\infty$ (and therefore does not exist in the strict sense)?
- Is the radius of convergence $R$ the whole $\mathbb{R}$ if $r'=\underset{n\to\infty}{\lim\operatorname{sup}}\left|\frac{c_n}{c_{n+1}}\right|$ is $\infty$? This is probably not true. What is a counterexample?
2 Answers
$\begingroup$The answer to question 1. is yes.
Let $R=|z-a|$. Since $\lim_{n\to\infty}\left|\frac{c_n}{c_{n+1}}\right|=\infty$ there is some $N$ such that $\left|\frac{c_n}{c_{n+1}}\right| > 2R$ for all $n\geq N$. By induction $\left|\frac{c_N}{c_n}\right| > (2R)^{n-N}$.
Thus $$\sum_{n=N}^\infty |c_n (z-a)^n| = |c_N|\sum_{n=N}^\infty\left|\frac{c_n}{c_N}\right|R^{n} < |c_N|\sum_{n=N}^\infty (2R)^{N-n}R^{n} $$ $$= |c_N|R^{N}\sum_{n=0}^\infty 2^{-n} = 2|c_N|R^N$$ so the series converges absolutely.
For a concrete counterexample for 2. take $c_n = \begin{cases} 1 & n\mbox{ even} \\ \frac{1}{n} & n\mbox{odd}\end{cases}$ and note that the series clearly diverges for $|z-a|\geq1$.
$\endgroup$ $\begingroup$Good question. The answer to 1 is affirmative. For 2, you really need the $\liminf$ to be infinite, which of course implies that the limit exists as $\infty$. This is relevant for finite radii of convergence. More precisely, d'Alembert's 1768 version of the ratio test, as stated on page 70 of Stromberg, states
Let $\sum c_n$ be a series of complex terms with $c_n \neq 0$ for all $n$.
- If $\limsup_{n\rightarrow\infty} \left|c_{n+1}/c_n\right|<1$, then the series converges absolutely.
- If $\liminf_{n\rightarrow\infty} \left|c_{n+1}/c_n\right|>1$, then hte series diverges.
Upon taking reciprocals in part 1, we see that we need the $\liminf$ for both sides.
To construct a specific counter example to your question, just take $c_{2n}=n^n$ and $c_{2n-1}=n^{-n}$. Then your $\limsup$ is infinite but the series converges nowhere except zero. In fact, you can construct just about any behavior you want using this even/odd alternating definitions.
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